Hello, TYTY!
With absolute values, a sketch usually helps . . .
$\displaystyle \int^2_{2}\left(13x\right)\,dx$
We know the graph of $\displaystyle y = x$ . . . a V with vertex at the origin.
The graph of $\displaystyle y = 3x$ . . . "climbs steeper".
The graph of $\displaystyle y \:=\:3x$ . . . inverts the previous graph.
The graph of $\displaystyle y \:=\:13x$ . . . moved up one unit. Code:

*
/\
/::\
2 /::::\ 2
  +  o  +  o  +  
::/  \::
:/  \:
/  \
*  *

We have an inverted V, vertex at (0,1), xintercepts at $\displaystyle \pm\tfrac{1}{3}$
They ask for the "area" from $\displaystyle x = \text{}2$ to $\displaystyle x=2,$
. . and some of it is "negative area".
We don't need Calculus to find the areas of triangles.
[My answer is: 8]